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  • The tangent at a point P on the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfra

The tangent at a point P on the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} passes through the point (0,-b) and the normal at P passes through the point \mathrm{(2 a \sqrt{2}, 0)}. Then the square of eccentricity of the hyperbola is

Option: 1

5


Option: 2

2


Option: 3

9


Option: 4

8


Answers (1)

best_answer

The equation of tangent at \mathrm{\left(x_1, y_1\right)} is

\mathrm{\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1}

It passes through (0,-b). So,

\mathrm{0+\frac{y_1}{b}=1 \text { or } y_1=b}

The equation of normal is

\mathrm{\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2 e^2 }
which passes through (2 a \sqrt{2}, 0). Hence,

\mathrm{\frac{a^2 2 a \sqrt{2}}{x_1}=a^2 c^2 \text { or } x_1=\frac{2 a \sqrt{2}}{c^2}
Now, \mathrm{\left(x_1, y_1\right)} lies on the hyperbola. Therefore,

\mathrm{\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1 \text { or } \frac{8}{e^4}-1=1 \text { or } e^2=2 }

Posted by

Irshad Anwar

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